- (a) Given that \(y = \ln \left[ t + \sqrt { } \left( 1 + t ^ { 2 } \right) \right]\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} t } = \frac { 1 } { \sqrt { } \left( 1 + t ^ { 2 } \right) }\).
The curve \(C\) has parametric equations
$$x = \frac { 1 } { \sqrt { } \left( 1 + t ^ { 2 } \right) } , \quad y = \ln \left[ t + \sqrt { } \left( 1 + t ^ { 2 } \right) \right] , \quad t \in \mathbb { R }$$
A student was asked to prove that, for \(t > 0\), the gradient of the tangent to \(C\) is negative.
The attempted proof was as follows:
$$\begin{aligned}
y & = \ln \left( t + \frac { 1 } { x } \right)
& = \ln \left( \frac { t x + 1 } { x } \right)
& = \ln ( t x + 1 ) - \ln x
\therefore \frac { \mathrm {~d} y } { \mathrm {~d} x } & = \frac { t } { t x + 1 } - \frac { 1 } { x }
& = \frac { \frac { t } { x } } { t + \frac { 1 } { x } } - \frac { 1 } { x }
& = \frac { t \sqrt { } \left( 1 + t ^ { 2 } \right) } { t + \sqrt { } \left( 1 + t ^ { 2 } \right) } - \sqrt { } \left( 1 + t ^ { 2 } \right)
& = - \frac { \left( 1 + t ^ { 2 } \right) } { t + \sqrt { } \left( 1 + t ^ { 2 } \right) }
\end{aligned}$$
As \(\left( 1 + t ^ { 2 } \right) > 0\), and \(t + \sqrt { } \left( 1 + t ^ { 2 } \right) > 0\) for \(t > 0 , \frac { \mathrm {~d} y } { \mathrm {~d} x } < 0\) for \(t > 0\).
(b) (i) Identify the error in this attempt.
(ii) Give a correct version of the proof.
(c) Prove that \(\ln \left[ - t + \sqrt { } \left( 1 + t ^ { 2 } \right) \right] = - \ln \left[ t + \sqrt { } \left( 1 + t ^ { 2 } \right) \right]\).
(d) Deduce that \(C\) is symmetric about the \(x\)-axis and sketch the graph of \(C\).